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[math-fun] ISC: 0; ries: 1
by Bill Gosper 06 Jun '18

06 Jun '18

Given 1/2 + 1/(2 + π), https://isc.carma.newcastle.edu.au/advancedCalc "really found nothing." ries took 4msec: Your target value: T = 0.694492264824171 mrob.com/ries . . . x-1/2 = 1/(2+pi) for x = T - 1.11022e-16 {78} --LHS-- --RHS-- -Total- max complexity: 44 38 82 dead-ends: 9401 11053 20454 Time: 0.004 expressions: 516 441 957 distinct: 444 347 791 Memory: 61136 B Total equations tested: 154068 (1.541e+05) But ISC knows scads more exotic constants and functions. --rwg Now my only problem is that .6944... is the difference between two quantities I thought were equal.

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[math-fun] Packing parallelograms into a triangle
by James Propp 05 Jun '18

05 Jun '18

Puzzle: Show that a triangle cannot be written as the union of a finite number of parallelograms with disjoint interiors. Puzzle: Show that you can cover n/(n+1) of the area of a triangle using n parallelograms with disjoint interiors. Problem: is it possible to do better? Jim Propp

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[math-fun] Sec@n/n →0??
by Bill Gosper 05 Jun '18

05 Jun '18

Abs exceeds 9 for four of the first thousand convergents of π/2: [62]:= Select[Sec[#]/# & /@ Numerator@Convergents[π/2, 999], Abs[#] > 9 &] //tim During evaluation of In[62]:= 0.213593,4 Out[62]= {Sec[11]/11, Sec[214112296674652]/214112296674652, Sec[1428599129020608582548671]/1428599129020608582548671, Sec[63008132762960627316194351129]/63008132762960627316194351129} And no others among the next hundred thousand! In[63]:= Select[Sec[#]/# & /@ Numerator@Convergents[π/2, 99999], Abs[#] > 9 &] //tim During evaluation of In[63]:= 25.228223,4 Out[63]= {Sec[11]/11, Sec[214112296674652]/214112296674652, Sec[1428599129020608582548671]/1428599129020608582548671, Sec[63008132762960627316194351129]/63008132762960627316194351129} !?! In[64]:= N[N[%, 33], 9] Out[64]= {20.5411872, 18.0078002, 11.5073819, 16.8247791} And none negative. Wouldn't Gauss's Log[2,1+tail] probability predict a fairly steady, non-tapering stream of qualifiers? --rwg On Sat, Jun 2, 2018 at 12:00 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > Out[42]= Sec[18190586279576483]/18190586279576483 > > In[43]:= {N@%, N[%, 22]} > > Out[43]= {-6.53302256185329*10^-17, 2.154106494388454341277} > --rwg > Does |Sec@n/n| (integer n) have a supremum, or even a simple max? > Hopelessly difficult: The minimal p for which Sec@n/n^p has a limit. >

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[math-fun] Testing 2, 3, 9, ...
by Henry Baker 04 Jun '18

04 Jun '18

This is just a test. If this had been an actual email, you would have been instructed to switch to ASCII from EBCDIC.

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Re: [math-fun] Leyland primes
by Keith F. Lynch 04 Jun '18

04 Jun '18

Hans Havermann recently mentioned Leyland primes (primes in the form x^y + y^x), so I looked them up on OEIS. I expected to see a bunch of random-looking odd numbers. And for the most part, that's what I saw. But I noticed that some of them have remarkably long runs of 0s. For instance the 30th term is a 465-digit decimal number, of which 305 consecutive digits are 0. And the 37th term is a 613-digit number of which 417 consecutive digits are 0. Why should that be? And does something similar happen in other radices? Thanks. One possibility is that in those cases y^x is much greater than x^y and y is divisible by 10. If so, I'd expect that I'd find primes in the form x^y - y^x to often have long runs of 9s. I found a list at A123206, and see that that is indeed the case. So I guess that's the explanation.

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[math-fun] Parametrization of the sphube
by françois mendzina essomba2 03 Jun '18

03 Jun '18

The points of the sphube of radius of axis r and center the origin of the mark can be parameterized by: x=r*(qos(theta))*(qos(phi)); y=r*(qos(theta))*(qin(phi)); z=r*qin(theta); theta in [sigma/2,-sigma/2]; Phi in [sigma,-sigma]; for the sphube x ^ 4 + y ^ 4 + z ^ 4 = 1 we have: r = 1

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[math-fun] On the existence of Leyland primes with a given number of decimal digits
by Hans Havermann 03 Jun '18

03 Jun '18

I've been searching (some ten months now) for Leyland prime numbers with about 100000 decimal digits. Leyland numbers are of the form x^y + y^x, x,y integers, x>=y>1. Primes of this form, of this size, are infrequent but I've got enough of them now to state that (empirically) the chances of there existing an n-digit Leyland prime for a given n ~ 100000 are about 1 in 75. Currently the largest known Leyland prime has 386434 decimal digits, so I got to wondering what the odds might be for the existence of an n-digit Leyland prime with n ~ 400000. My best guess is that they are about 1 in 95 but my math is somewhat contrived and I'm not at all sure of the result. Anybody want to have a go at this?

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Re: [math-fun] An application of chaos?
by Dan Asimov 02 Jun '18

02 Jun '18

A result in the same vein was that the zeta function zeta(x+iy) = Sum_{1 <= n < oo} 1/n^(x+iy) (defined, e.g., as above for x+iy with x > 1 and by analytic continuation everywhere else, except for x+iy = 1, where it's infinite). is sufficiently crazy (if you go far enough away from the x-axis on the critical strip given by {x+iy | 0 < x < 1}) that zeta(x+iy approximates any preassigned analytic function f(x+iy) *somewhere* to any degree of accuracy: (https://en.wikipedia.org/wiki/Riemann_zeta_function#Universality) —Dan >An interesting but probably useless result. > >https://marginalrevolution.com/marginalrevolution/2018/05/one-parameter-equ… > >Brent > >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] An application of chaos?
by Brent Meeker 02 Jun '18

02 Jun '18

An interesting but probably useless result. https://marginalrevolution.com/marginalrevolution/2018/05/one-parameter-equ… Brent

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Re: [math-fun] "I don't find this proof beautiful" quote?
by Bill Gosper 02 Jun '18

02 Jun '18

On 2018-05-29 09:59, Mike Stay wrote: > My favorite irrationality proof is one I heard from John Baez. > Suppose the cube root of two were not irrational; then there would be > two positive integers p, q such that p/q = ∛2. Multiplying both sides > by q and cubing, we get p³ = 2q³ = q³ + q³, which has no solutions in > the positive integers by Fermat's Last Theorem! I just read this aloud to Rohan, who remarked "You can also use that for ∛9.", barely looking up from his video game. --rwg

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